A 200 g block connected to a light spring with a force constant of k = 4 N/m osc

Question

A 200 g block connected to a light spring with a force constant of k = 4 N/m oscillates on a horizontal, frictionless surface. The block is released from rest with an initial displacement x_i = 4 cm from equilibrium. (A) What is the maximum magnitude of the force exerted by the spring during this motion? (B) At what value of x is the force from the spring equal to P = 38% of the maximum force? (C) At what value of x is the speed of the block equal to 38% of its maximum speed? The block is pulled back to stretch the spring, and is then released. As the block accelerates back and forth, the spring alternatively compresses and stretches, with maximum deformation at the locations where the block momentarily stops to reverse direction, and maximum speed where the block passes through the equilibrium point x = 0. Somewhere between the extremes are locations where the magnitude of the force has the specified value and other locations where the speed attains its specified value. | F | = (4 N/m)(4.00 times 10^-2 m) | - kx_max | = (4 N/m)(4.00 times 10^-2 m) = At what value of x is the force from the spring equal to 38% of the maximum force? Set | F | = | -kx | equal to 38% of the maximum value, without keeping track of the minus sign because the question asks about the magnitude of the force: From this it is seen that the force is reduced to 38% of its maximum magnitude where |x| has 38% of its maximum value, or at x = plusminus At what value of x is the speed of the block equal to 38% of its maximum speed? We can answer this question from the solutions for the position and velocity as functions of time. Because the block is at its maximum displacement x = x_0 at time t = 0, the equation for x is: with phi = 0. Therefore the equation for the velocity is: (2) v = -omega x_0 sin(omega t) whose maximum magnitude is omega x_0. The time r when the speed is 38% of this maximum value therefore satisfies: This is satisfied for: But because sin^2(omega t) + cos^2(omega t) = 1, Equation (4) reduces to: And therefore the locations were the speed has the required values are:

Explanation / Answer

Yes you will doing is correct but some small setups varaition

Maximum force =kx=4×4×10^-2=0.16N

(B)The value of X is equal 38% of maximum of force

Maximum force F=kx

If system contains maximun force =38/100 F

From above two equations are equal we get

38/100 F = kx

(38/100)0.16=4x

The x value at 38% of force =38×0.16/400

=0.0152m

=1.52 cm

(C) and

The block is at maximum displacement is X=Xo

At time t=0 so

The equation for X= Xo cos( wt)

Therefore the equation of velocity

V=Vo sin(wt)

Whose maximum magnitude Vo= wXo the time t whose speed is 38% of maximum value

Therefore it satisfying

(38/100)wXo =wXo sin (wt)

Therefore

Sin(wt) = 38/100

Because sin^2(wt)+cos^2(wt)=1

Using this trigonometry equation we get cos value

Cos(wt)=1-sin^2(wt)

=1-(38/100)^2

=0.925

And therefore the locations were has speed required value is

X=Xo cos(wt)

Here Xo is initial displacement at t=0 is 4cm

X =4×0.925

=3.7 cm

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.